[Leetcode]Nim Game

本文介绍了一种Nim游戏的策略分析方法,该游戏中两名玩家轮流从一堆石子中取走1到3个石子,取走最后一个石子的人获胜。文章通过分析得出,当石子总数除以4的余数不为0时,先手玩家可以赢得比赛。

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You are playing the following Nim Game with your friend: There is a heap of stones on the table, each time one of you take turns to remove 1 to 3 stones. The one who removes the last stone will be the winner. You will take the first turn to remove the stones.

Both of you are very clever and have optimal strategies for the game. Write a function to determine whether you can win the game given the number of stones in the heap.

For example, if there are 4 stones in the heap, then you will never win the game: no matter 1, 2, or 3 stones you remove, the last stone will always be removed by your friend.
此题的意思是一个游戏:桌上有一堆石子,我们每次只能拿1-3个,我先拿,最后拿石子的那个是赢家,我第一个拿。怎么样我才能赢。
最开始我在想用n%3,

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return (n%3 != n//3)

但是12的时候就不行,还是得4可以
用4,我就想了一下,应该这么写

class Solution(object):
    def canWinNim(self, n):
        """
        :type n: int
        :rtype: bool
        """
        return (n%4 != 0)         
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